Cute exponentials

Let $e^x = a$ , $e^y=b$ and $e^z=c$ . Then we have

$\begin{cases} ab + bc + ca = 1 &...(1) \\ a^2+b^2+c^2 = \dfrac {26}{27} + (abc)^2 &...(2) \end{cases}$

By Cauchy-Schwarz inequality on $(1)^2$ :

$\begin{aligned} (ab + bc + ca)^2 & \le \left(a^2+b^2+c^2\right) \left(b^2+c^2+a^2\right) \\ ab + bc + ca & \le a^2+b^2+c^2 \\ \implies a^2+b^2+c^2 & \ge 1 \end{aligned}$

$\implies$ The LHS of $(2) \ge 1$ .

By AM-GM inequality on $(1)$ :

$\begin{aligned} ab + bc + ca & \ge 3 \sqrt [3]{(abc)^2} \\ 1^3 & \ge 27 (abc)^2 \\ \implies (abc)^2 & \le \frac 1{27} \end{aligned}$

$\implies$ The RHS of $(2) = \dfrac {26}{27} + (abc) \le 1$ .

Since the LHS of $(2) \ge 1$ and the RHS of $(2) \le 1$ , LHS $=$ RHS, when the equality occurs to $3 \sqrt [3]{(abc)^2} \le ab+bc+ca \le a^2+b^2+c^2 = 1$ , when $a=b=c = \dfrac 1{\sqrt 3}$ .

$\begin{aligned} \implies e^{x+y+z} & = abc = \frac 1{3\sqrt 3} \\ \implies x+y + z & = \ln \left(\frac 1{3\sqrt 3} \right) = - \frac 32 \ln 3 \end{aligned}$

$\implies a+b + 1 = 2+3 + 1=\boxed{6}$